What are the solutions to the equation solvefor a,(a+b)^3=a^3+b^3 ?

The solutions to the equation solvefor a,(a+b)^3=a^3+b^3 are a=0,a=-b;\quad b\ne 0

Find the zeros of solvefor a,(a+b)^3=a^3+b^3

The zeros of solvefor a,(a+b)^3=a^3+b^3 are a=0,a=-b;\quad b\ne 0

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solve for

a, (a + b)^{3} = a^{3} + b^{3}

a = 0, a = - b; b \neq = 0

Solution steps

(a + b)^{3} = a^{3} + b^{3}

Expand

(a + b)^{3} : a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = a^{3} + b^{3}

Move

b^{3}

to the left side

a^{3} + 3 a^{2} b + 3 a b^{2} = a^{3}

Move

a^{3}

to the left side

3 a^{2} b + 3 a b^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

3 b a^{2} + 3 b^{2} a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 3 b ^{2} \pm ( 3 b ^{2} ) ^{2} - 4 \cdot 3 b \cdot 0}{2 \cdot 3 b}

Simplify

(3 b^{2})^{2} - 4 \cdot 3 b \cdot 0 : 3 b^{2}

a_{1, 2} = \frac{- 3 b ^{2} \pm 3 b ^{2}}{2 \cdot 3 b}; b \neq = 0

Separate the solutions

a_{1} = \frac{- 3 b ^{2} + 3 b ^{2}}{2 \cdot 3 b}, a_{2} = \frac{- 3 b ^{2} - 3 b ^{2}}{2 \cdot 3 b}

a = \frac{- 3 b ^{2} + 3 b ^{2}}{2 \cdot 3 b} : 0

a = \frac{- 3 b ^{2} - 3 b ^{2}}{2 \cdot 3 b} : - b

The solutions to the quadratic equation are:

a = 0, a = - b; b \neq = 0

x, x^{2} - 5 k x - 50 k (x)^{2} = 0

(m - 2) x^{2} - 2 m x + 3 m = 0

x = - 6 x^{2} + 8 x + 4

x^{2} - 3 x + \frac{1}{4} = 0

y, 2 x + 2 x y = 2 y + x^{2}

What are the solutions to the equation solvefor a,(a+b)^3=a^3+b^3 ?
The solutions to the equation solvefor a,(a+b)^3=a^3+b^3 are a=0,a=-b;\quad b\ne 0
Find the zeros of solvefor a,(a+b)^3=a^3+b^3
The zeros of solvefor a,(a+b)^3=a^3+b^3 are a=0,a=-b;\quad b\ne 0